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294 lines
9.9 KiB
Python
Executable file
294 lines
9.9 KiB
Python
Executable file
"""#
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### 谜题描述
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It's well known that the best way to distract from something is to do one's favourite thing. Job is such a thing for Leha.
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So the hacker began to work hard in order to get rid of boredom. It means that Leha began to hack computers all over the world. For such zeal boss gave the hacker a vacation of exactly x days. You know the majority of people prefer to go somewhere for a vacation, so Leha immediately went to the travel agency. There he found out that n vouchers left. i-th voucher is characterized by three integers li, ri, costi — day of departure from Vičkopolis, day of arriving back in Vičkopolis and cost of the voucher correspondingly. The duration of the i-th voucher is a value ri - li + 1.
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At the same time Leha wants to split his own vocation into two parts. Besides he wants to spend as little money as possible. Formally Leha wants to choose exactly two vouchers i and j (i ≠ j) so that they don't intersect, sum of their durations is exactly x and their total cost is as minimal as possible. Two vouchers i and j don't intersect if only at least one of the following conditions is fulfilled: ri < lj or rj < li.
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Help Leha to choose the necessary vouchers!
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Input
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The first line contains two integers n and x (2 ≤ n, x ≤ 2·105) — the number of vouchers in the travel agency and the duration of Leha's vacation correspondingly.
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Each of the next n lines contains three integers li, ri and costi (1 ≤ li ≤ ri ≤ 2·105, 1 ≤ costi ≤ 109) — description of the voucher.
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Output
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Print a single integer — a minimal amount of money that Leha will spend, or print - 1 if it's impossible to choose two disjoint vouchers with the total duration exactly x.
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Examples
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Input
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4 5
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1 3 4
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1 2 5
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5 6 1
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1 2 4
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Output
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5
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Input
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3 2
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4 6 3
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2 4 1
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3 5 4
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Output
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-1
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Note
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In the first sample Leha should choose first and third vouchers. Hereupon the total duration will be equal to (3 - 1 + 1) + (6 - 5 + 1) = 5 and the total cost will be 4 + 1 = 5.
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In the second sample the duration of each voucher is 3 therefore it's impossible to choose two vouchers with the total duration equal to 2.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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from collections import defaultdict
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from operator import itemgetter as it
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def build(seg_tree, n):
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for i in reversed(xrange(1, n)):
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if 2 * i > n - 1:
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seg_tree[i] = min(seg_tree[2 * i][2], seg_tree[2 * i + 1][2])
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else:
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seg_tree[i] = min(seg_tree[2 * i], seg_tree[2 * i + 1])
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def query(seg_tree, l, r, n):
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#l and r inclusive
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res = float('inf')
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l, r = l + n, r + n
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while l <= r:
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#if l is odd then its parent is not included so we have to include this when visiting left borders
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if l % 2 != 0:
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if l > n - 1: res = min(res, seg_tree[l][2])
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else: res = min(res, seg_tree[l])
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l += 1
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#if r is even then its parent is not included so we have to include this when visiting right borders
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if r % 2 == 0:
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if r > n - 1: res = min(res, seg_tree[r][2])
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else: res = min(res, seg_tree[r])
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r -= 1
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if l == r:
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break
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l, r = l / 2, r / 2
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return res
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def bs_floor(arr, val):
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l = -1
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r = len(arr)
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# -1 to n - 1
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while r - l > 1:
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mid = (l + r) / 2
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if arr[mid][0] <= val:
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l = mid
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else:
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r = mid
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return l
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n, x = map(int, raw_input().split())
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d = defaultdict(list)
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d1 = {}
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ans = float('inf')
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for a0 in xrange(1, n + 1):
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l, r, c = map(int, raw_input().split())
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d[r - l + 1].append((l, r, c))
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for key in d:
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d[key].sort(key=it(0))
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m = len(d[key])
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d1[key] = [0] * m + d[key][:]
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build(d1[key], m)
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for key in d:
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for (l, r, c) in d[key]:
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key1 = x - (r - l + 1)
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if key1 in d1:
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p = bs_floor(d[key1], r) + 1
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m = len(d[key1])
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if p != m:
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c1 = query(d1[key1], p, len(d[key1]) - 1, len(d[key1]))
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ans = min(ans, c + c1)
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if ans == float('inf'):
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print -1
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else:
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print ans
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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from collections import defaultdict
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import random
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import re
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import bisect
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class Chackerpackyourbagsbootcamp(Basebootcamp):
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def __init__(self, max_n=10, max_x=20, max_duration=200000, max_cost=10**9):
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self.max_n = max_n
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self.max_x = max_x
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self.max_duration = max_duration
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self.max_cost = max_cost
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def case_generator(self):
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x = random.randint(2, self.max_x)
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vouchers = []
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has_solution = random.choice([True, False])
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if has_solution:
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# 确保存在唯一最优解
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duration1 = random.randint(1, x-1)
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duration2 = x - duration1
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attempts = 0
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while True:
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# 生成第一个优惠券
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l1 = random.randint(1, self.max_duration - duration1)
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r1 = l1 + duration1 - 1
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# 生成第二个优惠券的位置
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if random.choice([True, False]):
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# 在第一个之后
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l2_min = r1 + 1
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r2_max = min(l2_min + duration2 + 3, self.max_duration)
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if r2_max - l2_min + 1 < duration2:
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continue
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l2 = random.randint(l2_min, r2_max - duration2 + 1)
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r2 = l2 + duration2 - 1
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else:
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# 在第一个之前
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r2_max = l1 - 1
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if r2_max < 1:
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continue
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l2_min = max(1, r2_max - duration2 + 1)
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r2 = random.randint(l2_min + duration2 - 1, r2_max)
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l2 = r2 - duration2 + 1
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# 验证区间有效性
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if (1 <= l2 <= r2 <= self.max_duration and
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(r1 < l2 or r2 < l1) and
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(r2 - l2 + 1) == duration2):
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break
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attempts += 1
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if attempts > 100:
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has_solution = False
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break
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if has_solution:
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# 生成最优解对
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c1 = random.randint(1, self.max_cost//4)
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c2 = random.randint(1, self.max_cost//4)
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vouchers = [(l1, r1, c1), (l2, l2 + duration2 - 1, c2)]
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# 添加干扰数据
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for _ in range(random.randint(0, self.max_n-2)):
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while True:
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l = random.randint(1, self.max_duration)
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r = random.randint(l, self.max_duration)
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d = r - l + 1
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if d + duration1 != x and d + duration2 != x:
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break
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cost = random.randint(max(c1, c2)+1, self.max_cost)
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vouchers.append((l, r, cost))
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if not has_solution:
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# 生成无解情况
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base_duration = x + 1
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min_duration = max(1, x//2 + 1)
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for _ in range(random.randint(2, self.max_n)):
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duration = random.randint(min_duration, x + 5)
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l = random.randint(1, self.max_duration - duration)
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r = l + duration - 1
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cost = random.randint(1, self.max_cost)
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vouchers.append((l, r, cost))
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random.shuffle(vouchers)
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return {
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'n': len(vouchers),
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'x': x,
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'vouchers': [{'l': l, 'r': r, 'cost': c} for (l, r, c) in vouchers]
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}
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@staticmethod
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def prompt_func(question_case):
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voucher_lines = '\n'.join(
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f"{v['l']} {v['r']} {v['cost']}"
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for v in question_case['vouchers']
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)
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return f"""Select two NON-OVERLAPPING travel vouchers meeting:
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1. Sum of durations EXACTLY {question_case['x']} days
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2. Minimum total cost
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Available vouchers:
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{voucher_lines}
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Output format: [answer]number[/answer] where number is the minimal cost or -1"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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try:
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return int(matches[-1].strip())
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except:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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vouchers = [(v['l'], v['r'], v['cost']) for v in identity['vouchers']]
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return solution == cls.optimal_solution(vouchers, identity['x'])
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@staticmethod
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def optimal_solution(vouchers, x):
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duration_map = defaultdict(list)
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for l, r, cost in vouchers:
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duration = r - l + 1
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duration_map[duration].append((l, cost))
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# 预处理每个duration的最小成本
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pre_min = {}
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for d in duration_map:
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sorted_list = sorted(duration_map[d], key=lambda x: x[0])
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min_prefix = []
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current_min = float('inf')
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for l, c in reversed(sorted_list):
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current_min = min(current_min, c)
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min_prefix.append(current_min)
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min_prefix.reverse()
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pre_min[d] = (sorted_list, min_prefix)
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min_cost = float('inf')
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for l, r, cost in vouchers:
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current_duration = r - l + 1
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need_duration = x - current_duration
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if need_duration not in pre_min:
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continue
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sorted_list, min_prefix = pre_min[need_duration]
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idx = bisect.bisect_right(sorted_list, (r, float('inf')))
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if idx < len(min_prefix):
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min_cost = min(min_cost, cost + min_prefix[idx])
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return min_cost if min_cost != float('inf') else -1
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